A chemist reduces 10.0 mmol of methyl 4-oxopentanoate, which has one ketone group and one ester…

Step 1. Ketone 1, ester 2: $1 + 2 = 3$ hydride ions for each molecule.

Step 2. Total hydride ions: $10.0 \times 3 = 30.0\ \text{mmol}$. Each $\text{LiAlH}_{4}$ gives 4, so $\dfrac{30.0}{4} = 7.50\ \text{mmol}$.

Step 3. Mass: $7.50 \times 10^{-3}\ \text{mol} \times 37.95\ \text{g/mol} = 0.285\ \text{g}$.

Step 4. The product. Both groups are reduced: the ketone to a secondary alcohol, and the ester to a primary alcohol and methanol. The product is pentane-1,4-diol.